Forum:Last digits of tree(3)
TREE(3) is an enormous number, proven to be even larger than Graham's Number. While there is no way to calculate the last decimal digits, it is possible to calculate the last binary digits. This can be done by recording the total number of tree "branches" on every odd number in binary digits. (We start with 0, not 1) —Preceding unsigned comment added by 107.77.161.10 (talk • ) 10:59, September 17, 2017 (UTC) :The problem with trying to calculate TREE(3) is that it is the maximal sequence for TREE(3) is not given by any specific procedure; rather, it is the longest sequence that satisfies certain requirements, so it is hard to imagine how you could find it other than to search over all possible such sequences. :If you indeed have a way to calculate the last binary digits, could you please explain further? You haven't given nearly enough for us to understand what you are talking about. :I will say that I would be willing to bet that TREE(3) is odd, as the same for any TREE(n). This will be true if the maximum sequence is a sequence of nonpaths followed by a sequence of paths, or if it is a sequence of trees of height greater than 1 followed by a sequence of trees of height 1. (Both ending on the root tree, of course) The problem is, I don't know how one would prove that we can't have a path or a tree of height 1 come up earlier in the sequence. (It may be possible.) Deedlit11 (talk) 23:54, September 20, 2017 (UTC) :Here is how to do it: :Let n be the "branch height", or how far down the branch goes. Also, let d be the number of digits, or the number of nonnegative odd numbers at or below n. Starting with n = 1 and any odd value n, we begin this process: :1. We use the binary number 11 and keep the last digit (1) because d = 1. :2. Next, we go down to n = 3. Here, we have the number of dots at 10'11'. Now, we keep 11 and those are now that last binary digits. :3. To further prove my point, let's change n to 5. Now, we have the number 111''' 11'''. :4. Continue the pattern if you want. :That's how we calculate the last binary digits of TREE(3) —Preceding unsigned comment added by 222.97.145.127 (talk • ) 00:03, October 12, 2017 (UTC) ::The branch height of what? We don't know what the specific trees are in the longest possible sequence. Also, you say "let d be the number of digits, or the number of nonnegative odd numbers at or below n". Which one is it? Those are two different things. Then you say "Starting with n = 1 and any odd value n" - is n equal to 1 or any odd value? Please be more clear. Deedlit11 (talk) 00:25, October 12, 2017 (UTC) :: Here is a link for a picture to explain it: http://108.50.218.20/tree3digits.jpg —Preceding unsigned comment added by 107.77.161.12 (talk • ) 19:25, October 12, 2017 (UTC) ::User Deedlit11 just said that he wants to know what the branch height is. and what I mean "let d be the number of digits, or the number of nonnegative odd numbers at or below n". I mean with the next question with "Starting with n = 1 and any odd value n" that n is equal to any odd value, including 1. I will reiterate what I have said before in a more clear and thought-out way (I'm re-explaining it almost from scratch): ::1. Use the argument as the variable d and accept any d > 0. ::2. Assign the variable n for n branches or sticks down as (2d + 1). (the branches equal the height of the tree.) ::3. Go down the tree to a height of n (the tree we are using is n high). ::4. Count the number of points or connectors (the dots). ::5. Convert the number of points/dots to binary. ::6. Only use the last d digits. ::7. Those are your last binary digits of TREE(3). ::—Preceding unsigned comment added by 222.97.145.127 (talk • ) 22:08, October 13, 2017 (UTC) ::WaxPlanck (talk) 13:44, October 28, 2017 (UTC) The other user didn't understand how to do this, but here is how the right way works: ::1. Start with 3 dots. (we'll call them "seeds".) ::2. Make the dots red, green, and blue. (This gives you the digit of 1.) ::3. Make the red dot have nothing on it. (it dies.) ::4. Draw a line from the green dot down to the green dot. (It dies after that.) ::5. Now, draw a red and green dot down from the blue dot. (this doesn't count as a provable digit, all of the branches in this "forest" after this will.) ::6. Draw 3 lines down from the red with red, green, and blue and 4 lines down from the green with red, green, blue, and red. (This gives you the digits 11.) ::7. Draw more and more down from each branch, each branch being 1 more tree than the last. ::8. Keep going, repeating step 7. ::NOTE: If your tree has the same start and end colors, this means that the tree DIES.) ::(By using this method, I managed to compute the last 3 binary digits: 011.) :::How do you know that this method gives you the digits of TREE(3)? LittlePeng9 (talk) 14:07, October 28, 2017 (UTC) :::WaxPlanck (talk) 15:51, October 28, 2017 (UTC) Because those digits repeat later.